The electric field produced by a single point charge is given by:
[tex]E(r) = k \frac{q}{r^2} [/tex]
where
k is the Coulomb's constant
q is the charge
r is the distance from the charge
In our problem, the charge is [tex]q=7.2 \cdot 10^{-6} C[/tex], while the distance is [tex]r=1.6 m[/tex], therefore the intensity of the electric field at that distance is
[tex]E=k \frac{q}{r^2}=(8.99 \cdot 10^{9} Nm^2C^{-2}) \frac{(7.2 \cdot 10^{-6} C)}{(1.6m)^2} =2.53 \cdot 10^4 N/C[/tex]
and since the charge is positive, the electric field is directed away from the charge (so, eastwards)
